What is the approximate standard free energy | vedclass

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(A) For a Daniell cell,the cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$The standard Gibbs free energy chang

Vedclass · Accepted Answer

$-212.3$

Vedclass · Answer

(A) For a Daniell cell,the cell reaction is: $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$The standard Gibbs free energy change is given by the formula: $\Delta G^{\circ} = -nFE^{\circ}_{cell}$Here,$n = 2$ (number of electrons transferred),$F = 96500 \ C \ mol^{-1}$ (Faraday's constant),$E^{\circ}_{cell} = 1.1 \ V$ (standard emf of the Daniell cell).Substituting these values:$\Delta G^{\circ} = -2 	imes 96500 \ C \ mol^{-1} 	imes 1.1 \ V$$\Delta G^{\circ} = -212300 \ J \ mol^{-1}$Converting to $kJ \ mol^{-1}$:$\Delta G^{\circ} = -212.3 \ kJ \ mol^{-1}$

Electrolyte	$\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$
$NaCl$	$126.5$
$NaBr$	$128.2$
$KCl$	$149.5$

What is the approximate standard free energy change per mole of $Zn$ (in $kJ \ mol^{-1}$) for a Daniell cell at $298 \ K$?

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Electrolyte $\Lambda_{m}^{o} \, (S \, cm^{2} \, mol^{-1})$

$NaCl$ $126.5$

$NaBr$ $128.2$

$KCl$ $149.5$

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